# Example 12.1
# Determine (a) Field circuit resistance (b) Field rheostat setting that will
# provide no load voltage of 140V (c) Armature voltage if the rheostat is set
# to 14.23 ohm (d) Field rheostat setting that will cause critical resistance
# (e) Armature voltage at 80 percent rated speed (f) Rheostat setting required
# to obtain no load armature voltage of 140V if shunt field is separately
# excited from a 120V DC source
# Page No. 479
# Given data
Ea=156.; # No load voltage
If=4.7; # Shunt field current
If140=2.35; # New field current at Ea=140V
Eanew=140; # No load voltage
Ifnew=3.2; # Field current corresponding to no load voltage
Ea1=0; # First arbitrary voltage
Ea2=100.; # Second arbitrary voltage
Vf=120.;
V=130.; # Intersection of I1 and I2
Rrheonew=14.42; # Rheostat set to new settings
Va=116.; # Intersection of field resistance line with the low
# speed magnetization curve
# (a) Field circuit resistance
Rf=Ea/If; # Field circuit resistance
# (b) Field rheostat setting that will provide no load voltage of 140V
Rrheo=(Eanew/Ifnew)-Rf;
# (c) Armature voltage if the rheostat is set to 14.23 ohm
Rnew=Rf+Rrheonew; # New field resistance
If1=Ea1/(Rf+Rrheo); # Field current corresponding to first arbitrary voltage
If2=Ea2/(Rf+Rrheo); # Field current corresponding to second arbitrary voltage
# (d) Field rheostat setting that will cause critical resistance
Rcr=Eanew/If140; # Critical resistance
# (e) Armature voltage at 80 percent rated speed
# Ea80=0.80*Ea;
Ea80=116;
# (f) Rheostat setting required to obtain no load armature voltage of 140V if
# shunt field is separately excited from a 120V DC source
Rrheo1=(Vf/Ifnew)-Rf;
# Display result on command window
print"Field circuit resistance =",Rf,"Ohm"
print"Field rheostat setting that will provide no load voltage of 140V =",Rrheo,"Ohm"
print"Armature voltage if the rheostat is set to 14.23 ohm =",V,"V"
print"Field rheostat setting that will cause critical resistance =",Rcr,"Ohm"
print"Armature voltage at 80 percent rated speed (V)=",Ea80
print"Rheostat setting required =",Rrheo1,"Ohm"
# Example 12.2
# Computation of (a) No load voltage (b) Voltage regulation
# (c) Resistance setting of rheostat necessary to obtain rated voltage
# at rated conditions
# Page No. 487
# Given data
P=300000.; # Shunt generator power rating
VT=240.; # Shunt generator voltage rating
Ra=0.00234; # Armature winding resistance
RIP=0.00080; # Resistance of interpole winding
Fnet=5100.; # Net mmf
Vnl=255.; # No load voltage
Vrated=240.; # Rated voltage
Nf=1020.; # Turns per pole
Vf=120.; # Source that separately excites the generator
If=5.69;
Rf=18.1;
# (a) No load voltage
Ia=P/VT; # Armature current
Ea=VT+Ia*(Ra+RIP); # Armature emf
Ff=Fnet/(1.-0.121);
# (b) Voltage regulation
VR=(Vnl-Vrated)*100./Vrated;
# (c) Resistance setting of rheostat necessary to obtain rated voltage at rated conditions
If=Ff/Nf;
Rrheo=(Vf/If)-Rf; # Rheostat setting
# Display result on command window
print"No load voltage =",Vnl,"V"
print"Voltage regulation =",VR,"Percent"
print"Resistance setting of rheostat necessary =",Rrheo,"Ohm"
# Example 12.3
# Computation of (a) Induced emf at rated load (b) No load voltage
# (c) Voltage regulation (d) What is the type of compounding?
# Page No. 492
# Given data
Pload=320000.; # Shunt generator power rating
Vrated=250.; # Shunt generator voltage rating
Rf=20.2; # Shunt resistance
Rrheo=7.70; # Shunt field rheostat value
If=8.96; # Field current
Iload=1280.; # Load current
Ra=0.00817; # Armature resistance
Rip=0.00238; # Resistance of interpole winding
Rse=0.00109; # Resistance of series winding
Nf=502.; # Turns per pole
VNL=225.; # No load voltage
# (a) Induced emf at rated load
Iload=Pload/Vrated; # Load current
If=Vrated/(Rf+Rrheo); # Field current
Ia=If+Iload; # Armature current
Racir=Ra+Rip+Rse;
Ea=Vrated+Ia*Racir;
# (b) No load voltage
Ff=Nf*If;
# (c) Voltage regulation
VR=(VNL-Vrated)*100./Vrated;
# Display result on command window
print"Induced emf at rated load =",Ea,"V"
print"No load voltage =",VNL,"V"
print"Voltage regulation =",VR,"Percent"
print"The machine is overcompounded"
# Example 12.4
# Computation of (a) Required resistance of a noninductive diverter that will
# bypass 27 percent of the total armature current(b) Power rating of the
# diverter
# Page No. 494
# Given data
Rs=0.00306; # Shunt generator resistance rating
Is=0.73; # Shunt generator current rating
Id1=0.27; # Armature winding resistance
Pload=170000.; # Load of power
VT=250.; # Shunt generator voltage rating
Id2=680.; # No load voltage
Rd=0.27; # Resistance drop
# (a) Required resistance of a noninductive diverter that will bypass
# 27 percent of the total armature current
Rd=Rs*Is/Id1;
# (b) Power rating of the diverter
Ia=Pload/VT;
Pd=((Id1*Id2)**2.)*Rd;
# Display result on command window
print"Required resistance of a noninductive diverter = %0.5f Ohm ",Rd
print"Power rating of the diverter =",Pd,"W "
# Example 12.5
# Computation of (a) New bus voltage (b) Current supplied by each generator
# Page No. 500
# Given data
p1=300000.; # Rated power in generator A
p2=400000.; # Rated power in generator B
v=250.; # Rated voltage in machine
p3=350000.; # Rated power in generator C
Ibnew=2500.;
# (a) New bus voltage
IArated=p1/v; # Rated current in generator A
IBrated=p2/v; # Rated current in generator B
IBorig=p3/v; # Original bus current
IbDelta=Ibnew-IBorig; # Current difference
DelVbus=IbDelta/(160.+128.); # Voltage difference
# (b) Current supplied by each generator
DelIA=160.*DelVbus; # Generator A current difference
DelIB=128.*DelVbus; # Generator A current difference
Vbus=v-DelVbus; # Voltage across the bus
IA=700.+DelIA; # Current in generator A
IB=700.+DelIB; # Current in generator B
Loading= (IA-IArated)*100./IArated;
# Display result on command window
print"New bus voltage =",DelVbus,"V"
print"Current supplied by generator A =",IA,"A"
print"Current supplied by generator B =",IB,"A"
print"Macine A is overloaded by",Loading,"Percent"
# Example 12.6
# Determine (a) The increment increase in load on each machine if an
# additional 400 A load is connected to the bus (b) Current carried
# by each machine
# Page No. 502
# Given data
p1=100000.; # Rated power in generator A
p2=300000.; # Rated power in generator B
v=250.; # Rated voltage in machine
p3=30000.; # Rated power in generator C
Ibnew=400.; # New bus current
I1=200.;
I2=500.;
# (a) The increment increase in load on each machine if an additional 400 A
# load is connected to the bus
IArated=p1/v; # Rated current in generator A
IBrated=p2/v; # Rated current in generator B
Ib=p3/v; # Original bus current
DelVbus=Ibnew/(40.+120.); # Change in bus current
DelIA=40.*DelVbus;
DelIB=120.*DelVbus;
# (b) Current carried by each machine
IA=I1+DelIA; # Current in generator A
IB=I2+DelIB; # Current in generator B
# Display result on command window
print"The increment increase in load on machine A =",DelIA,"A"
print"The increment increase in load on machine B =",DelIB,"A"
print"Current carried by machine A =",IA,"A"
print"Current carried by machine B =",IB,"A"